Question

22. Predict the major product for the following reactions. OsO4 CH3 CH3COOH н* / H50 H2O2 HBr / peroxide 1. Hg(OAc)2, THF, CH3OH 2. NaBH4, HO

Answer 1

anti-product syn-product JOH 1111111 -CH3 CH2COOH CHz CH3 OsO4 H2O2 H/H20 HBr/peroxide 1. Hg(OAc)2, THF, CH3OH 2. NaBH4, OH mercuration reaction OME On CH3 CH3 Br anti-Markovnikov product Note: If instead of CH3OH, water is there then alcohol is formed in the oxymercuration reaction. This reaction gives a Markovnikov product. (Note: The hydrogen will go to the more hindered carbone)

1. (A) Alkyne is considered to be 2 elements of unsaturation.

(B) One ring is considered as 1 element of unsaturation and there are two rings that mean, there are also 2 elements of unsaturation.

(C) This compound contains 1 ring and 1 double bond which suggests this also has 2 elements of unsaturation.

(D) This compound also contains 1 ring and 1 double bond which suggests this also has 2 elements of unsaturation.

Hence, compound A-D all contain 2 elements of unsaturation.

Correct Answer is (E) all of these

2.

H3CO HO Br / (Z)-3,7-dibromo-5-(1-chloroethyl)-4-methoxymethyl)- 6-methylhept-4-en-1-ol

Correct Answer is (B) Z  

3. Bredt's rule says that a double bond can't be present at the bridgehead position in bicyclic compounds. This condition is relaxed if the bridgehead double bond is present in more than a six-membered cyclic ring.

In the case of (A), although the bridgehead double bond is present but this is not a bicyclic compound. Hence, not violating Bredt's rule.

In the case of (B), there is a bridgehead double bond in a bicyclic ring of the compound. Hence, it is violating the Bredt's rule.

In the case of (C), no bridgehead double bond so not violating Bredt's rule.

In the case of (D), although the bridgehead double bond present but it is a 8-membered ring that relaxed the Bredt's rule.

Hence, the correct Answer is (B)

4. Hence, the correct answer is (A).

2. 6 2,6-dimethyl-1,3,5-octatriene

5. We know that the compound which contains the same group on one of the carbons of a double bond will not show the E-Z isomerism.

Here, in the four options, option number (C) has two methyl groups on one of the carbon. That means it will not show E-Z isomerism.

Hence, the correct answer is (C)