5
$\begingroup$

According to me the poisoned catalyst will only reduce the triple bond so the compound produced must be optically active.. But the answer given is that the final product is optically inactive.. is it correct and if so how?

(2Z,4S)-4-methylhept-2-en-5-yne

this is the final product I made:

(2Z,5Z)-4-methylhepta-2,5-diene

The triple bond gets reduced to double bond as Lindlar's catalyst has been used so the reduction stops at this stage.. is this correct?

$\endgroup$
4
  • $\begingroup$ What do you think you get when you reduce the triple bond? Have you drawn the structure? $\endgroup$
    – matt_black
    Feb 12 '17 at 23:54
  • $\begingroup$ I have posted the structure I made.. Please check it @matt $\endgroup$ Feb 13 '17 at 0:00
  • 2
    $\begingroup$ And is this optically active? Does it have a carbon with 4 distinct units attached? $\endgroup$
    – matt_black
    Feb 13 '17 at 0:09
  • 1
    $\begingroup$ sorry.. I overlooked the 2 group[s are same.. Thanks.. i understood.. $\endgroup$ Feb 13 '17 at 0:14
7
$\begingroup$

The two cis-configured pro-1-en-1-yl groups present on the central carbon remove the asymmetry. You can put a plane of symmetry through the $\ce{Me-C-H}$ plane of the central carbon to transform the molecule onto itself. You can confirm that by rotating the $\ce{(C=)C-C(HMeR)}$ bond.

Thus, this hydrogenation removes chirality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.