Question

Draw structural formulas for the major product(s) formed by reaction of 3 -hexyne with each of these reagents. (Where you predict no reaction, write NR.) (a) \(\mathrm{H}_{2}\) (excess)/ \(\mathrm{Pt}\) (b) \(\mathrm{H}_{2} /\) Lindlar catalyst (c) \(\mathrm{Na}\) in \(\mathrm{NH}_{3}(l)\) (d) \(\mathrm{BH}_{3}\) followed by \(\mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{NaOH}\) (e) \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOH}\) (f) \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOD}\) (g) \(\mathrm{Cl}_{2}(1 \mathrm{~mol})\) (h) \(\mathrm{NaNH}_{2}\) in \(\mathrm{NH}_{3}(l)\) (i) \(\mathrm{HBr}(1 \mathrm{~mol})\) (j) \(\mathrm{HBr}(2 \mathrm{~mol})\) (k) \(\mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{HgSO}_{4}\)

Short answer

a) H2/Pt b) H2/Lindlar catalyst c) Na in NH3(l) d) BH3 followed by H2O2/NaOH e) BH3 followed by CH3COOH f) BH3 followed by CH3COOD g) Cl2 (1 mol) h) NaNH2 in NH3(l) i) HBr (1 mol) j) HBr (2 mol) k) H2O in H2SO4/HgSO4 Answer: a) 3-hexane b) 3-hexene (cis-alkene) c) (E)-3-hexene (trans-alkene) d) 3-hexanol (anti-Markovnikov alcohol) e) NR (No Reaction) f) NR (No Reaction) g) 3,4-dichlorohex-1-yne h) 3-hexynide anion (conjugate base) i) 3-bromohex-1-yne j) 3,4-dibromohexane k) 2-hexanone

Step-by-step solution

(a) Reaction with \(\mathrm{H}_{2}\)/ \(\mathrm{Pt}\)

Hydrogenation takes place under these conditions. In the presence of \(\mathrm{Pt}\) and excess hydrogen, the 3-hexyne molecule undergoes complete hydrogenation to form the alkane 3-hexane.

(b) Reaction with \(\mathrm{H}_{2}\)/ Lindlar catalyst

This is partial hydrogenation. With the Lindlar catalyst, the 3-hexyne molecule undergoes partial hydrogenation to form the cis-alkene, which is 3-hexene.

(c) Reaction with \(\mathrm{Na}\) in \(\mathrm{NH}_{3}(l)\)

This reaction is a metal-ammonia reduction. The 3-hexyne undergoes reduction to form the trans-alkene, which is (E)-3-hexene.

(d) Reaction with \(\mathrm{BH}_{3}\) followed by \(\mathrm{H}_{2} \mathrm{O}_{2} / \mathrm{NaOH}\)

This reaction is known as hydroboration-oxidation. The 3-hexyne reacts to form the anti-Markovnikov alcohol, which is 3-hexanol.

(e) Reaction with \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOH}\)

No reaction occurs here, so we write NR.

(f) Reaction with \(\mathrm{BH}_{3}\) followed by \(\mathrm{CH}_{3} \mathrm{COOD}\)

No reaction occurs here, so we write NR.

(g) Reaction with \(\mathrm{Cl}_{2}(1 \mathrm{~mol})\)

This is a halogenation reaction. The 3-hexyne reacts with 1 mol of \(\mathrm{Cl}_{2}\) to produce 3,4-dichlorohex-1-yne.

(h) Reaction with \(\mathrm{NaNH}_{2}\) in \(\mathrm{NH}_{3}(l)\)

This reaction is a deprotonation reaction. The 3-hexyne loses a proton to form a conjugate base, which is the 3-hexynide anion.

(i) Reaction with \(\mathrm{HBr}(1 \mathrm{~mol})\)

This is a hydrohalogenation reaction. The 3-hexyne reacts with 1 mol of \(\mathrm{HBr}\) to produce 3-bromohex-1-yne.

(j) Reaction with \(\mathrm{HBr}(2 \mathrm{~mol})\)

In the presence of 2 moles of \(\mathrm{HBr}\), 3-hexyne undergoes a double hydrohalogenation, which results in the formation of 3,4-dibromohexane.

(k) Reaction with \(\mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{HgSO}_{4}\)

This is a hydration reaction. The 3-hexyne reacts to form a ketone, which is then converted to 2-hexanone.