1 1 butinin asagdakilerin her biriyle reaksiyona girdiginde...

1. 1-Bütin'in aşağıdakilerin her biriyle reaksiyona girdiğinde oluşmasını beklediğiniz ürünleri reaksiyon mekanizması basamaklarını göstererek yazınız. a) Alumina varlığında 1 eşdeğer mol HBr ardından benzoil peroksit ve 1 eşdeğer mol HBr b) Sıvı NH3 içerisinde NaNH2 ardından etil iyodür 2. Aşağıda verilen bileşiğin sentezi için iki ayrı yöntem öneriniz ve bu yöntemleri mekanizmaları ile gösteriniz. 3. 3,3-dimetil bütan-2-ol den başlayarak 2,3-dimetil, 2-bütanol sentezini mekanizması ile gösteriniz. 4. Başlangıç maddesi olarak en fazla dört karbonlu bir alkolden başlayarak 2,5-dimetilhekzan-3-on un sentezini tasarlayınız. 2,5-dimetilhekzan-3-on 5. Bir mol asetilenden başlayarak 2 mol asetik asit sentezi için bir yöntem öneriniz ve reaksiyon mekanizmaları ile gösteriniz.

Transcript

00:01 Hello students, this question consists of multiple part.

00:04 In the first part we are going to prepare ethyl 2 -methyl -3 -oxopentanoate from ethyl propanoate.

00:11 This is the ethyl propanoate.

00:13 This ethyl propanoate will undergo glycine condensation.

00:16 The first step is the proton abstraction.

00:19 The proton is abstracted by the ethoxide ion and this has been eliminated as ethanol and this result in the formation of a negative charge over here.

00:30 This negative charge will attack the carbonyl carbon of another molecule of ethyl propanoate.

00:34 Therefore, this bond will shits over here and this result in the formation of a negative charge over the oxygen that can be written as follows ch2o, c double bond o and this has been attached to a ch and a methyl group and this has been attached to o and ch2, ch3 and here will be the negative charged oxygen and here will be our ethyl group.

01:00 Now, this negative charge will comes over here and this ethoxide ion will be eliminated and this result in the formation of our expected product that is ethyl group, oxygen and a carbonyl group and another carbonyl group and here will be a methyl group and here will be the ethyl substituent.

01:24 So, this is the answer to the first question.

01:26 Coming to the second question, michael reaction of ethyl acetoacetate with buta -3 -ene -2 -one.

01:33 Michael reaction is nothing but the michael addition.

01:36 In the first step, the hydrogen has been abstracted by the hydroxide ion.

01:42 So, this result in the formation of a negative charge over here.

01:45 This negative charge will attack this carbonyl carbon and this will shits over here and this will shits over here and result in the formation of our intermediate with a negative charge over the oxygen and double bond o.

02:01 This has been attached to double bond o and here will be the second compound and here will be our oxygen group with o-.

02:12 This o - will abstract the hydrogen from the water molecule and thus hydroxide ion will be eliminated and thus result in the formation of a enol compound.

02:25 So, c double bond och double bond bond oh.

02:32 This hydrogen will be eliminated and this bond shits over here and this bond which shits over here and the eliminated hydrogen will attack over here...

Question

Please give Ace some feedback