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Predict the products of the following reactions:

  1. ²ú³Ü³Ù²¹²Ô-2-´Ç²Ô±ð â¶Ä‰+ â¶Ä‰d¾±±ð³Ù³ó²â±ô²¹³¾¾±²Ô±ð→NaBH(OAc)3
  2. 4-fluoropyridine→NaOCH2CH3
  3. 3-nitroaniline→(2)CuBr(1)HCl,NaNO2
  4. butan-2-one→(2)LiAlH4(1)KCN,HCN
  5. cyclopentanone→(2)LiAlH4(1)aniline,H+
  6. 2-bromopentane→(2)Ag2O,heat(1)(CH3)3N:

h.

i.

j.

Short Answer

Expert verified

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Step by step solution

01

Prediction of products

(a) Butan-2-one reacts with diethylamine in presence of sodium triacetoxyborohydride to give the final product as N,N-diethylbutan-2-amine.

reaction a

(b) 4-fluoropyridine on treatment with sodium ethoxide gives the product as 4-ethoxypyridine.

reaction b

(c) 3-nitroaniline on treatment with hydrochloric acid and sodium nitrite followed by copper bromide gives the final product as 3-bromoaniline.

reaction c

(d) Butan-2-one on treatment with potassium cyanide and hydrogen cyanide gives the initial product as 2-hydroxy-2-methylbutanenitrile, which on further eduction with lithium aluminum hydride gives the final product as 1-amino-2-methylbutan-2-ol.

reaction d

(e) Cyclopentanone reacts with aniline in presence of an acid to give the initial product as N-cyclopentylideneaniline, , which on further eduction with lithium aluminum hydride gives the final product as N-cyclopentylaniline.

reaction e

(f) 2-bromopetane on treatment with trimethylaniline gives the initial product as N,N,N-trimethylpentan-2-aminium, which on further treatement with silver oxide followed by heating gives the final products as pent-1-ene and trimethylamine.

reaction f

(g) Cyclohexanone on treatment with ethylamine in presence of an acid followed by the loss of a water molecule gives the product as N-cyclohexylideneethanamine.

reaction g

(h) Cyclohexanone on treatment with diethylamine followed by reduction with sodium triacetoxyborohydride gives the product as N-ethylcyclohexanamine.

reaction h

(i) The given compound on treatment with hydrogen peroxide followed by heating gives the product as 3,3-dimethylcyclohex-1-ene.

reaction i

(j) The given compound on treatment with excess methyl iodide followed by another treatment with silver oxide and heating gives the product as as shown below.

reaction j

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Most popular questions from this chapter

The two most general amine syntheses are the reductive amination of carbonyl compounds and the reduction of amides. Show how these techniques can be used to accomplish the following syntheses.

(a) benzoic acid to benzylamine

(b) benzaldehyde to benzylamine

(c) pyrrolidine to N-ethylpyrrolidine

(d) cyclohexanone to N- cyclohexylpyrrolidine

(e) HOOC-(CH2)3 -COOH to pentane-1,5-diamine cadaverine

Addition of one equivalent of ammonia to 1-bromoheptane gives a mixture of heptan-1-amine, some dialkylamine, some trialkylamine, and even some tetraalkylammonium bromide.

  1. Give a mechanism to show how this reaction takes place, as far as the dialkylamine.
  2. How would you modify the procedure to get an acceptable yield of heptan-1-amine?

Reductive amination of aldehydes and ketones is a versatile method for attaching alkyl groups to amines, but the alkyl group is restricted to a 1 0or 20carbon by this method. Prof. Phil Baran of Scripps Research Institute has reported (Science, 2015, 348(6237), 886-891) a novel way to reduce an aromatic nitro group and add the resulting amine to an alkene so that the aromatic amine is bonded to a carbon- all in a continuous sequence of reactions. For example:

Predict the products using these starting materials, all of which are reported in this paper.

Rank the amines in each set in order of increasing basicity.

a) Guanidine (shown) is about as strong base as hydroxide ion. Explain why guanidine is a much stronger base than most other amines.

(b) Show why p-nitroaniline is a much weaker base (3 pKb units weaker) than aniline.

(c) Explain why N,N-2,6-tetramethylaniline (shown) is a much stronger base than N,N-dimethylaniline.

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