Question

b) When the following deuterated compound reacts under the same conditions, the rate of formation of the substitution product is unchanged, but the elimination rate is slowed by a factor of 7. Why is the elimination slowed, but substitution unchanged? D3C-CH-CD3 + KOH - D2C=CH-CD3 + D3C-CH-CD3 Br alcohol OH (20) 7. Which reaction of the following pairs will take place more rapidly? a) CH, CH, CH, Br + CH, OH — CH, CH, CH, Br + CH30 — b) CH. Br + CH2OH CH, Br + CH, SH - c) (CH3)2CHCH,CI (CH3)2CHCH, I + N3. - + N = d) - — (Polar Protic Solvents) CHBr + (CH3)3N CH, Br + (CH3)3P CHZI + OH- - CH, I + CHŞCO, e) —

Answer 1

6 when the deuterated en the deuterated Compound reacts under the same the rate of the formation of substitution product is unchanged but the climination is slowed by a bactor of I The elimination is slowed because of the fact that Capture of s by bare is not so much facile as compare to tH' - becaun of cd bond is more strengthen than C-H bond. "D3C- CH - CD3 + KOH - D2C=CH-CD3 QlCowo 20.4) eliminationbar. à than & D3-CH-CD3 OH (substitution Nucleophilicity Broduct) of a charged species is more than an uncharged one. So CH-CH2-CH-Port Chaus reaction will take place more rapidly an Ctz-Brit CH₃ SH will takes place more rapidly because of greolen nucleophilieity of Stow'o' . (et, CH-Cik-I the will takes place more rapidly because It is a good leaving group than a CH₃-Port (CH3)3 P-will taken blace more rapidly beansin olan brutic Solvent 'N' remains t' bonded-sunucleophilicity decreasy. (Hz It OHO) will taxes blare more rapidly became ructrophilicity of it is higher than to ao it remains as C

CH-CH2-ONa tetz-CHI - CH₂ - CH₂ - O-che-CH3 Hs por (CH3)2 – CH – CH3 + 8 cctty) (CH3) CH - CH - o -c (CH3) zalkyl haride - So together with the william etier Synthesis elimination also can happened. Son Son other possibility of product- etly ne=CH-CH3 (more substituted) HOGHS + CH₂ - CH - CI CH3 – CH₂ - CH2 - Hocults (Itz Hz-cit- & -o-GH5 -HT @ Question is wrong Уч, у chzucht- e - 0 - GH5 Citz Ⓡc-c6e - wts + Batuh cinta CH ROH CH₃ -CH2-O-CH2 - CH₂ good leaving group

гол гора, group linee Bensic to H VOUCH Ph am H ITCH CH3 -CH2-CHtz - Ph H + wcph ph (trans stilbenes (2) minor) emajun) So only 7 is formed

6. 10.) Hz (Hz (-o-ctly - The best Method to prepare this I would be the williamson ether synthesis where we have to us (atz), cogna and Clip-I Cetty), co + City I ~ (CH3)3 C-0 -CH₂ the other souteis (13)3 c I and CH₂ but asit is a sN2type reaction, the route would not be a correct one. Cits et-ut & CH₂ - - 1 ditz Sardou not occur due to steric factor so the above will be the better choice.