Question

How might the geometry of the product change (oh in an endo or exo position?) if all the methyl groups of camphor were replaced with h? 3. the reduction mechanism is often shown with a?

Answer 1

Replacement of methyl groups of camphor is a reduction mechanism. Camphor is a bridged bi-cyclic compound. If you observe the structure of camphor the methyl group is placed with one carbon bridge (7, 7) and two carbon bridge (1). Attack from face of one carbon bridge is termed as exo attack whereas from face of two carbon bridge is termed as endo attack. So replacement will lead to mix of both and formation of two stereo isomers.

Reduction mechanism is often shown with a U-shaped arrow pointing the attack by ion such as in nucleophilic addition reaction.

Answer 2

The replacement of all the methyl groups yields preferentially the product with the exo OH group.

Further Explanation:

Stereoisomerism is a form of isomerism in which molecules have the same molecular formula and sequence of bonded atoms. The difference is in the three-dimensional orientation of their atoms in space. It is also known as spatial isomerism.

Stereoisomerism is of mainly two types:  

1. Configurational stereoisomerism

2. Conformational stereoisomerism

Endo-Exo isomerism is a type of stereoisomerism that are found in organic compounds with a substituent on a bridged ring system. The prefix endo is used when the substituent is on the same side or syn of the longest bridge and the prefix Exo is used when the substituent is on the opposite side or anti of the longest bridge. Longest here refers to the bridge with the highest number of atoms. (Refer to the image attached below)

Camphor $\left( {{{\text{C}}_{{\text{10}}}}{{\text{H}}_{{\text{16}}}}{\text{O}}} \right)$ has a shorter bridge comprising of three carbon atoms, two longer bridge of four atoms, three methyl substituents and a carbonyl group.

Since the carbonyl group is ${\text{s}}{{\text{p}}^2}$ hybridized molecule so it is planar species. On its reduction, the reducing agent can approach the carbonyl carbon by above the plane of bond or below the plane of bond because the lobes of vacant p orbital are present above and below the plane of the carbonyl bond. So, two product stereoisomers are formed but there amounts depends on the stability of the product.

Borneol (exo OH group) is synthesized by exo attack of reducing agent but due to steric repulsion of two methyl groups on exo side, the reduction product formed is small and preferentially endo product is formed that is, Isoborneol(with endo OH group).

When all the methyl groups are replaced by hydrogen atoms compound is called norcamphor and its reduction yields preferentially the exo product(exo-norborneol) and endo product(endo-norborneol). The product formed from endo approach is more hindered unlikely to camphor. (Refer to the attached image)

Learn more:

1. Which ester will form on the reaction between propanoic acid and isopropyl alcohol: brainly.com/question/5902825

2. Draw every stereoisomer for 1,2-difluoro-1,2-dimethylcyclopentane: brainly.com/question/8803065

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Stereochemistry

Keywords: Endo, Exo, camphor, stereoisomer, methyl, OH, norcamphor and carbonyl group.