tag:blogger.com,1999:blog-89932813722410907952024-03-12T21:11:41.766-07:00Organic chemistryThis blog is a learning resource for undergraduate students studying organic chemistry. It consists of a database of actual questions and answers about organic chemistry collected by a chemistry professor teaching the subject. Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.comBlogger428125tag:blogger.com,1999:blog-8993281372241090795.post-36146899560114179072023-10-30T11:01:00.002-07:002023-10-30T12:19:04.504-07:00<p><span id="docs-internal-guid-45853b37-7fff-0e44-f7a7-4ca9b65cfd28"></span></p><p dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-left: 40.5pt; margin-top: 0pt; text-align: justify;"><span style="background-color: white; color: black; font-family: 'Times New Roman',serif; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Can you re-explain why nucleophilicity does not correlate with basicity under protic conditions? </span><span style="background-color: white; color: black; font-family: 'Times New Roman',serif; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Because the larger particle size </span><span style="background-color: white; color: black; font-family: 'Times New Roman',serif; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 700; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">decreases</span><span style="background-color: white; color: black; font-family: 'Times New Roman',serif; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;"> their basicity and </span><span style="background-color: white; color: black; font-family: 'Times New Roman',serif; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 700; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">nucleophilicity</span><span style="background-color: white; color: black; font-family: 'Times New Roman',serif; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;"> due to stabilization </span><span style="background-color: white; color: black; font-family: 'Times New Roman',serif; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 700; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">and increases nucleophilicity </span><span style="background-color: white; color: black; font-family: 'Times New Roman',serif; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">due to polarization. In protic solvents the basicity is suppressed, but nucleophilicity is not affected, so polarization overrides stabilization. Great question! I added it to our organic chemistry blog. - GS). </span></p><div><span style="background-color: white; font-family: "Times New Roman", serif; font-size: 12pt; font-style: italic; font-variant-alternates: normal; font-variant-east-asian: normal; font-variant-numeric: normal; font-variant-position: normal; text-align: justify; vertical-align: baseline; white-space-collapse: preserve;"><br /></span></div>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-85642278946255322512022-11-02T09:53:00.002-07:002022-11-02T09:55:56.467-07:00<p> <span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">Can you clarify the differences between nucleophilicity </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">(Nucleophiles </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-decoration-line: underline; text-decoration-skip-ink: none; vertical-align: baseline; white-space: pre-wrap;">change</span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;"> their ability to donate a pair of electrons in the process of the reaction due to polarizability. They usually attack carbon - GS).</span><span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">, basicity </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">(Bases </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-decoration-line: underline; text-decoration-skip-ink: none; vertical-align: baseline; white-space: pre-wrap;">do not change</span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;"> their ability to donate a pair of electrons in the process of the reaction. They usually attack carbon - GS)</span><span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">, and polarizability </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">(Ease of separation of charges in the electric field, which makes the difference between bases and nucleophiles - GS)</span></p>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-19461658445239104562022-10-05T18:42:00.000-07:002022-10-05T18:42:28.440-07:00Markovnikov rule<p> F<span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;">or which reactions do we use markovnikov’s rule versus anti-markovnikov’s rule? </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;">(Regardless of the rule, the preferred reaction proceeds through the most stable cation or radical on the most substituted carbon. It requires the least substituted carbon of the C=C bond to be attacked. So, when the attacking particle is hydrogen (such as H</span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;"><span style="font-size: 0.6em; vertical-align: super;">+</span></span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;"> in acid-catalyzed hydration or hydrohalogenation) or something that will become hydrogen later (Hg(OAc)</span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;"><span style="font-size: 0.6em; vertical-align: sub;">2</span></span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;"> in mercuration followed by reduction, the reaction goes by Markovnikov rule. If the least substituted carbon is attacked with something else (Br</span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;"><span style="font-size: 0.6em; vertical-align: super;">.</span></span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;"> in free radical hydrobromination) or Boron in hydroboration followed by oxidation, the reaction goes against Markovnikov rule - GS).</span></p><span id="docs-internal-guid-165210d4-7fff-b0e2-fa66-83c3b1650b54"><br /></span>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-8673995227917474832021-03-16T21:01:00.000-07:002021-03-16T21:02:12.059-07:00<p> <span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;">How can you tell when a reaction goes through the meisenheimer complex or benzyne? </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;">(The Meisenheimer complex is formed only in the presence of a resonance acceptor </span><span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;">in an ortho- or para-position to the leaving group. Otherwise the reaction requires a strong base, harsher conditions, and proceeds through benzyne - GS).</span></p><div><span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;"><br /></span></div>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-52053905917384721652020-10-22T20:33:00.000-07:002020-10-22T20:34:20.659-07:00<p> <span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;">Can you explain in more detail the separation of enantiomers into diastereomers? </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;">(A mixture of enantiomers is treated with a chiral reagent to convert them to two new compounds, which have opposite centers from enantiomers and same centers from the chiral reagent, so they are diastereomers. Next, the diastereomers are separated and cleaved back into original enantiomers and the chiral reagent back. - GS).</span></p><div><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; text-align: justify; vertical-align: baseline; white-space: pre-wrap;"><br /></span></div>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-69843280731690666392020-04-28T12:12:00.001-07:002020-04-28T12:12:18.934-07:00<span id="docs-internal-guid-e973c4e1-7fff-cf7e-9cc2-1233e33dad1d"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt; text-align: justify;">
<span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">What should we know regarding fluorescence? I recall discussing the topic in the past but am unable to find my notes regarding. </span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(When a molecule absorbs the energy of light, it can cause a chemical reaction (photochemistry), or re-emit back as a light of different energy (fluorescence), or “wiggle out” by molecular vibrations to heat. Rigidity of a molecule reduces “wiggling out” the energy, which increases the efficiency of fluorescence. Therefore, we can control fluorescence by affecting the molecule by pH, presence of metals, DNA, etc. That is how many chemical sensors work. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-41397952086340581082020-04-27T21:07:00.001-07:002020-04-27T21:08:16.346-07:00<span id="docs-internal-guid-55a820f3-7fff-a11e-8d17-10c328ab3b34"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt; text-align: justify;">
<span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">E1 reactions are not stereoselective, so would the products be present in a racemic mixture like Sn1? </span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(No. Unlike SN1, E1 does not produce a chiral center because the C=C bond is planar. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-25985877405414264302020-04-25T16:28:00.001-07:002020-04-25T16:28:26.397-07:00<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt; text-align: justify;">
<span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">So would the addition of acetic anhydride to an amine be acylation? </span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Yes. It is actually acyl nucleophilic substitution, because half of the anhydride molecule leaves. - GS).</span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;"> Also are there the same issues with acylation as there are with alkylation when trying to make a primary amine or not because acylation is substitution? </span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(No. The C=O group in an acyl blocks the lone electron pair on nitrogen from the second acylation. That is why acylation can be used as a get-around way to avoid the issue with multiple alkylations. Example - Gabriel synthesis. - GS).</span></div>
<span id="docs-internal-guid-27303497-7fff-e969-c9bb-7cd59dbead32"><br /><br /></span>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-30539744375632552662020-04-14T12:26:00.001-07:002020-04-14T12:26:19.009-07:00<span id="docs-internal-guid-1c20df1c-7fff-f976-e718-6977f367c4a4"><span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">Why would the hydrogens attached to the carbons in an ether and amine not be able to be hydrogen bond donors? </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">(The small difference in electronegativity between C and H leads to the low polarity of the C-H bond, which leads to a very small positive charge on H, so it does not interact with the lone electron pair of the H-bond acceptor efficiently. - GS).</span></span>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-75486441344218614772020-04-05T17:10:00.001-07:002020-04-05T17:10:19.845-07:00<span id="docs-internal-guid-f44792a3-7fff-8aa1-ceda-565f17a3bcc0"><span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">For ortho meta para directors, it is the group that is already attached to the ring that “tells” the group that is adding where to go or is it the group that is adding that knows where to add “on its own”? </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">(The group, which is already in the ring, because it participates in the stabilization of the intermediate sigma-complex. The incoming group is separated from the positively charged conjugated system of the sigma-complex by an sp3-carbon, so its influence is minimal.</span></span>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-5536784681817813852020-03-31T16:47:00.002-07:002020-03-31T16:47:23.420-07:00<span id="docs-internal-guid-1325fd8e-7fff-7d07-f621-d11fd2fdc048"><span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">When looking at strong bases resulting in E rxns why does the major product favor Hoffman if the LG is poor? Such as F. </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">(Reluctance of the leaving group to leave deviates the mechanism from the synchronous E2: Abstraction of a proton by a base leads to accumulation of negative charge on the carbon (in a classical E2, the developing negative charge is immediately carried away by the leaving group). As a consequence, the reaction pathway is determined by the stability of a carboanion. The trend of stability of carbanions is the opposite to that of carbocations, that is the carbanion is more stable at the less substituted carbon. Therefore, the less substituted C=C is produced, which is against Zaitsev rule. - GS).</span></span>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-63110593854719227432020-01-23T13:34:00.004-08:002020-01-23T13:34:40.098-08:00N-14 NMR<span id="docs-internal-guid-10009337-7fff-c530-0bbe-7d0e6f9c6eb0"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt; text-align: justify;">
<span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">How can 14^N be NMR active if it has 7 N and 7 Protons? I understood the text as both must be even numbers for the nucleus to be active (TA). </span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Not quite. If the number of neutrons and protons is odd, the nucleus is always magnetic, because the magnetic moments of individual nuclei can not compensate for each other. If this number is even, they usually do, and the nucleus is not magnetic. Sometimes, the magnetic moments of nuclei are aligned (two nuclei have the same magnetic moments, instead of the opposite moments), and then the nucleus is magnetic. </span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;"><span style="font-size: 0.6em; vertical-align: super;">14</span></span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">N is an example (so-called quadrupole nucleus). Another simpler example is deuterium (</span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;"><span style="font-size: 0.6em; vertical-align: super;">2</span></span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">H) - GS. </span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-61659676419861093362018-03-26T12:00:00.001-07:002018-03-26T12:00:53.021-07:00<span id="docs-internal-guid-f03c2492-63ae-d958-6cf2-1c39894e9a40"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt; text-align: justify;">
<span style="background-color: transparent; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Why are SN2 reactions better in aprotic? </span><span style="background-color: transparent; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Poor solvation of the nucleophile favors SN2, while poor solvation of the leaving group disfavors SN1. - GS).</span><span style="background-color: transparent; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;"> Are both Sn2 and Sn1 reactions better in polar vs nonpolar? </span><span style="background-color: transparent; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Yes. Polarity helps formation of a carbocation (for SN1) and helps to dissolve the nucleophile (for both SN2 and SN1). - GS). </span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-59780356483136393062018-03-21T05:21:00.001-07:002018-03-21T05:21:24.324-07:00<span id="docs-internal-guid-f19854ec-4881-6647-a062-ed3f0e9be51c"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt; text-align: justify;">
<span style="background-color: transparent; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Can we please re-discuss why protic solvents cause products to form faster and why they are more favored than aprotic solvents in SN1 reactions? </span><span style="background-color: transparent; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Because protic solvents solvate anions in addition to carbocations. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-11188990625661190782018-03-13T19:00:00.001-07:002018-03-13T19:01:36.977-07:00<span id="docs-internal-guid-f19854ec-223c-cc5b-d81f-8306f3cbfeb2"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt; text-align: justify;">
<span style="background-color: transparent; color: black; font-family: "times new roman"; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre;">Can you explain conformation vs configuration? </span><span style="background-color: transparent; color: black; font-family: "times new roman"; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre;">(Configuration can be changed only by breaking chemical bonds. Conformation changes when a molecule adopts a different shape, without breaking bonds. - GS). </span><span style="background-color: transparent; color: black; font-family: "times new roman"; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre;">So E/Z would be considered a configuration? </span><span style="background-color: transparent; color: black; font-family: "times new roman"; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre;">(Yes. - GS). </span><span style="background-color: transparent; color: black; font-family: "times new roman"; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre;">Okay, thanks!</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-54005446375177443462018-03-13T15:05:00.001-07:002018-03-13T15:05:18.367-07:00<span id="docs-internal-guid-f19854ec-2165-2bb8-598d-c13a473d9b41"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">What is the difference between NaNh2 and Na, NH3 (This first reagent is a strong base, the second one is a strong reducing agent. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-55918295899055723022018-03-13T15:04:00.001-07:002018-03-13T15:04:46.905-07:00<span id="docs-internal-guid-f19854ec-2164-9928-9b5b-ba25239c44bf"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Why does halogenation of alkynes make stereoselectivity of addition unpredictable? </span><span style="background-color: white; color: black; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Because the 3-membered bromonium responsible for anti-stereoselectivity, is unstable (too strained) - GS). </span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-33386461314990839222018-03-13T15:03:00.001-07:002018-03-13T15:04:11.397-07:00<span id="docs-internal-guid-f19854ec-2162-f623-86f8-b51c6e263f92"><span style="background-color: white; font-size: 12pt; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">What are other ways to create an acetylenide other than using NaNH2? </span><span style="background-color: white; font-size: 12pt; font-style: italic; font-variant-east-asian: normal; font-variant-numeric: normal; vertical-align: baseline; white-space: pre-wrap;">(Grignard reagents, alkyllithium compounds, such as Butyl lithium - GS).</span></span>Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-50599168838064267852018-02-13T20:12:00.001-08:002018-02-13T20:13:02.190-08:00<span id="docs-internal-guid-89f89879-9283-cbdc-3451-f878032a8492"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Will you please explain the difference between cationic polymerization of alkenes and free radical polymerization of alkenes. And how that compares to addition of HCl to alkenes. </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Both cationic polymerization and electrophilic addition of HCl start from protonation of C=C. Radical polymerization starts from the attack of C=C by a free radical. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-68897669980419678542018-02-13T18:28:00.001-08:002018-02-13T18:28:55.645-08:00<span id="docs-internal-guid-89f89879-9224-7961-42f4-43d4d55446b1"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">I was reviewing through the lecture slides and came across the following question: </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 700; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Why do we not consider absorption of IR by N2 and O2, which are way more abundant in the atmosphere, then CO2?</span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;"> Is the answer because N2 and O2 are symmetrical, unlike CO2, and therefore they have a negligible dipole moment; and, with IR absorption such as in IR spectroscopy, a dipole moment must be present in order for IR to be absorbed? Thank you. </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Almost. We need to consider change of the dipole moment, not the dipole moment itself. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-91036517112180185482018-02-13T18:27:00.003-08:002018-02-13T18:27:55.897-08:00<span id="docs-internal-guid-89f89879-9223-aa9d-743d-4508883d8373"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Why is carbon 13 used instead of carbon 12 for NMR? </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Because Carbon 13 is magnetic, and carbon - 12 is not. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-75907557893156551192018-02-13T18:27:00.001-08:002018-02-13T18:27:25.774-08:00<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">In hydrobromination in the presence of peroxide, does the hydrogen or bromine connect to the carbon of the c=c? </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Both, just like in any addition to C=C. - GS).</span></div>
<span id="docs-internal-guid-89f89879-9223-325d-ee92-87cd8232d547"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Which one connects first, the hydrogen or bromine? </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Bromine. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-54216197856723728712018-02-13T18:26:00.001-08:002018-02-13T18:26:49.550-08:00<span id="docs-internal-guid-89f89879-9222-51fb-20c4-433dbf9187ba"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: #222222; font-family: "times new roman"; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre;">Do bromonium cations only form with cyclic compounds, or would it also form for straight chain alkenes? </span><span style="background-color: white; color: #222222; font-family: "times new roman"; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre;">(Any compound shape. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-88225299927452903492018-02-13T18:25:00.001-08:002018-02-13T18:25:44.436-08:00<span id="docs-internal-guid-89f89879-9221-6253-be41-32efba15abdf"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">How would the free-radical addition of Br-Br to cyclohexene compare to the non-free radical addition? </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Under free-radical conditions, Br2 will substitute an allylic hydrogen rather than add to C=C. NBS is even a better reagent for free radical halogenation. - GS). </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Why does it substitute an allylic hydrogen instead of adding to the C=C? </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Because abstraction of an allyl hydrogen leads to a stable allyl radical, so the reaction pathway shifts to the more favorable substitution over addition. Some addition still takes place, and this is suppressed by using NBS instead of Br2. - GS).</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0tag:blogger.com,1999:blog-8993281372241090795.post-83487632220194654372018-02-11T10:09:00.001-08:002018-02-11T10:09:41.817-08:00<span id="docs-internal-guid-89f89879-860e-0251-87f4-ca3cefc9ad49"></span><br />
<div dir="ltr" style="line-height: 1.38; margin-bottom: 0pt; margin-top: 0pt;">
<span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: normal; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">Should we be able to predict carbocation rearrangements, or just know how/why they occur? </span><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">(Yes., You can predict the rearrangement, if there is a pathway to a more stable carbocation by the following mechanism: </span><a href="http://ochem.orgfree.com/rear.html" style="text-decoration: none;"><span style="-webkit-text-decoration-skip: none; background-color: white; color: #1155cc; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration-skip-ink: none; text-decoration: underline; vertical-align: baseline; white-space: pre-wrap; white-space: pre;">http://ochem.orgfree.com/rear.html</span></a><span style="background-color: white; color: #222222; font-family: 'Times New Roman'; font-size: 12pt; font-style: italic; font-variant: normal; font-weight: 400; text-decoration: none; vertical-align: baseline; white-space: pre-wrap; white-space: pre;"> - GS.</span></div>
Grigoriy Seredahttp://www.blogger.com/profile/01642701864848937189noreply@blogger.com0