For each pair, predict the stronger nucleophile in the

Organic Chemistry | 8th Edition | ISBN: 9780321768414 | Authors: L. G. Wade Jr

Problem 16P Chapter 6

Organic Chemistry | 8th Edition

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Organic Chemistry | 8th Edition | ISBN: 9780321768414 | Authors: L. G. Wade Jr

Organic Chemistry | 8th Edition

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Problem 16P

For each pair, predict the stronger nucleophile in the reaction (using an alcohol as the solvent). Explain your prediction.

Step-by-Step Solution:

Solution 16P

Step 1 </p>

The  reaction relies on the rear-side attack of a strong nucleophile on a sterically unhindered electrophile containing a good leaving group.

(a)

The pair of compounds are  and.

Note that  is a secondary amine, and is less hindered than the other compound (which is a tertiary amine). There is less steric hindrance around the secondary amine, and so it will be the stronger nucleophile; it can more easily attack the electrophile.

Therefore,  will be the stronger nucleophile.

Step 2 </p>

(b)

The pair of compounds are  and.

Sulfur is a larger atom than oxygen, meaning that its outer valence electrons (the ones that participate in the  reaction) are further away from the positively charged nucleus, and therefore less tightly held. This makes the electrons more “polarizable”, and makes it easier for them to reach out and attack the electrophile.

Therefore  will be the stronger nucleophile.

Step 3 </p>

(c)

The pair of compounds are  and.

Phosphorus is a larger atom than nitrogen, meaning that its outer valence electrons (the ones that participate in the  reaction) are further away from the positively charged nucleus, and therefore less tightly held. This makes the electrons more “polarizable”, and makes it easier for them to reach out and attack the electrophile.

Therefore  will be the stronger nucleophile.

Step 4 </p>

(d)

The pair of compounds are  and.

Alkyl groups are mildly electron-donating via hyperconjugation effects, and anions of a given element are always stronger nucleophiles that neutral atoms of that element due to the extra buildup of negative charge.

Therefore  will be the stronger nucleophile of this pair.

Step 5 of 7

Chapter 6, Problem 16P is Solved
Step 6 of 7

Textbook: Organic Chemistry
Edition: 8
Author: L. G. Wade Jr
ISBN: 9780321768414

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