1,2-Dibromopropane on treatment with X moles of NaNH_2 followed by treatment with ethyl bromide gave a pentyne. The value of X is

1,2-Dibromopropane on treatment with X moles of NaNH_2 followed by tre

1.	1,2-Dibromopropane on treatment with X moles of NaNH_2 followed by treatment with ethyl bromide gave a pentyne. The value of X is
Answer» <html><body><p>One<<a href="https://interviewquestions.tuteehub.com/tag/br-390993" style="font-weight:bold;" target="_blank" title="Click to know more about BR">BR</a>>Two <br/><a href="https://interviewquestions.tuteehub.com/tag/three-708969" style="font-weight:bold;" target="_blank" title="Click to know more about THREE">THREE</a> <br/>Four </p>Solution :i) `CH_3CHBr - CH_2Br + 2NaNH_2 to CH_3C -= <a href="https://interviewquestions.tuteehub.com/tag/ch-913588" style="font-weight:bold;" target="_blank" title="Click to know more about CH">CH</a> + 2NaBr + 2NH_3` <br/> <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) `CH_3 C -= CH + NaNH_2 to CH_3C -= C^(-) Na^(+) + NH_3` <br/> iii) `CH_3C -= C^(-) Na^(+) + CH_3CH_2 Br to CH_3C -= C CH_2 CH_3 + NaBr` <br/> Thus 3 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of `NaNH_2` is used .</body></html>