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The halogenalkane in B is primary and C is tertiary. Primary halogenalkanes have SN-2 reactions which involve one step. Tertiary halogenalkanes have SN-1 reactions which involve two steps.
so shouldn't C be right?
C does have two steps but they have asked which doesn't so B is right as it doesn't.
I didn't count. but must be in the 20s somewhere.
I am not too sure about this but I think D is the right answer because as you down the alkane homologous series, the boiling point and melting point increases. So, maybe more energy would be needed for larger alkanes to undergo combustion and therefore more residual gas is produced by smaller alkanes over a certain period of time.
U dont even have 2 do any calculations of da stoichiometry. Valency of C is 4, valency of Al is 3, so u cross multiply n forms Al₄C₃. All other options wud fail 2 satisfy da octet rulemy doubt too
if u got the answer plz tag me
Question 26 is quite easy. Both W and Y would have to get changed to a halogenoalkane before the final products are formed.Please help with these...i will really appreciate it!
Ans B
1 and 2 correct
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