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Chapter 10: Q.33. (page 420)

Question: (a) Label the carbon–carbon double bond in A as E or Z. (b) Draw the products (including stereoisomers) formed when A is treated with H2O in the presence of H2SO4.

Short Answer

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Answer

(a) Z configuration

(b)

Step by step solution

01

Step 1:E or Z configuration

When the priority groups are on the same side of the double bond, it is called Z configuration, and when they are on different sides, it is called E configuration.

02

Formation of products for a and b

(a)

The given compound is depicted below:

Given compound

The configuration of the given compound is represented as follows:

Z configuration

(b)

The reaction of the given compound with water in the presence of sulfuric acid gives the stereoisomers of alcohol.

The reaction is shown below:

Products in b

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Most popular questions from this chapter

Question: Draw the structure corresponding to each IUPAC name.

a. (Z)-4-ethylhept-3-ene

b. (E)-3,5,6-trimethyloct-2-ene

c. (Z)-2-bromo-1-iodohex-1-ene

Question: How many degrees of unsaturation are present in each compound?

  1. C6H6
  2. role="math" localid="1648637368534" C8H18
  3. C7H8O
  4. C7H11Br
  5. C5H9N

Question: Draw the products formed when each alkene is treated with HCl.

Question: a. Which diastereomer of oct-4-ene yields a mixture of two enantiomers, (4R,5R)- and (4S,5S)-4,5-dibromooctane on reaction with Br2 ?

b. Which diastereomer of oct-4-ene yields a single meso compound, (4R,5S)-4,5-dibromooctane?

Question: Draw a stepwise mechanism for the following reaction, which results in ring expansion of a six-membered ring to a seven-membered ring.
See all solutions

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