Theoretically, butanoyl chloride can be synthesized from butane as starting material.

Question

Outline how this reaction can be completed requiring at least FIVE steps reactions. Draw

all relevant compounds (as starting material and product) and state the suitable reagent

Answer 1

Steps 1 & 2) Starting with the basic hydrolysis of the salt, sodium butanoate, C₃H₇CO₂Na, followed by acid hydrolysis using (say( HCl, we get butanoic acid.

1) C₃H₇CO₂Na + OH^- → C₃H₇CO₂^-

2 C₃H₇CO₂^- + H⁺ → C₃H₇CO₂H

Step 3) React the butanoic acid with methanol, using and acid to catalyze the reaction forming the ester, methyl butanoate.

3) C3H₇CO₂H + CH₃OH {cat of H⁺) → C₃H₇CO₂CH₃ + H₂O

Step 4) Hydrolyze the ester back to butanoic acid.

4) C₃H₇CO₂CH₃ + H₂O (cat of H⁺) → C₃H₇CO₂H + CH₃OH

Step 5) Reaction of the butanoic acid with a halogenating agent (such as SOCl₂ or PCl₃.

4) C₃H₇CO₂H + PCl₃ → C₃H₇COCl

It is acknowledged that this is rather repetitive & cumbersome and can be accomplished in only 4 steps, but the question wanted "at least" 5 steps, so.....

Answer 2

From butane...

1) Radical bromination (Br₂ in presence of light) of butane will yield 2-bromobutane as the major product. Recall that radical bromination will give preference for the higher degree alkyl bromide (in this case, secondary over primary).

CH₃CH₂CH₂CH₃ + Br₂ (+ light/heat) --------> CH₃CH₂CHBrCH₃ + HBr

2) Elimination of 2-bromobutane to yield 1-butene. To do this, you need a strong, bulky base such as potassium tert-butoxide, K^{+ -}OC(CH₃)₃. Recall that using a strong, bulky base will lead to formation of the less substituted alkene, i.e. the Hoffman product, as the preferred product. From a mechanistic point of view this reaction pathway follows the E2 mechanism.

CH₃CH₂CHBrCH₃ + K^{+ -}OC(CH₃)₃ -------> CH₃CH₂CH=CH₂ + HOC(CH₃)₃ + KBr

3) Hydroboration-oxidation of 1-butene to yield 1-butanol. The object here is to hydrate the alkene but in an anti-Markovnikov fashion, i.e. incorporation of the hydroxyl on the less substituted carbon of the starting alkene. The reaction conditions are: 1) B₂H₆; 2) HOOH, NaOH, H₂O; 3) H₃O⁺ (acid workup)

CH₃CH₂CH=CH₂ --------> CH₃CH₂CH₂CH₂OH

4) Chromic acid oxidation of 1-butanol to butanoic acid. Recall that a primary alcohol can be oxidized all the way to the carboxylic acid with chromic acid. Understand that the primary alcohol does get oxidized initially to the aldehyde but under the reaction conditions (H₂CrO₄, acetone, water, H₂SO₄), the aldehyde rapidly oxidizes further to the acid.

CH₃CH₂CH₂CH₂OH --------> CH₃CH₂CH₂COOH (butanoic acid)

5) Conversion of butanoic acid to butanoyl chloride using thionyl chloride, SOCl₂.

CH₃CH₂CH₂COOH + SOCl₂ ---------> CH₃CH₂CH₂COCl + HCl + SO₂

Hope this helps!!