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atroni [7]
3 years ago
7

Draw the structure corresponding to the common name α−chloro−β−methylbutyric acid

Chemistry
2 answers:
erica [24]3 years ago
8 0

Explanation:

α−chloro−β−methylbutyric acid  : 2-chloro 3-methyl butanoic acid

The main carbon chain in the α−chloro−β−methylbutyric acid  is of four carbon atom.Prefix 'buta' means four carbon atom are present in given carboxylic acid chain.

From on of the end one carbon is bonded to two oxygen and serves as carboxylic group in the molecule.

Chlorine atom is present on the second carbon (α-carbon)atom from the side where carbon is bonded to two oxygen atoms.

Methyl group  is present on the third carbon atom (β-carbon) from the side where carbon is bonded to two oxygen atoms.

Structure is given in an image attached.

uysha [10]3 years ago
6 0

Structure of a-chloro-b-methylbutyric acid (2-chloro-3-methylbutyric acid is attached below.


The structure was drawn using following information.


1) First, the parent chain was identified. The parent chain is containing four carbon atoms, also, this chain belongs to carboxylic acid as the compound name ends with Butyric acid i.e. butyr (Butane) -ic acid (carboxylic acid).


2) Secondly, the positions of substituents were assigned by using the rule which sates that the numbering in carboxylic acid must satrt from the carbonyl carbon. Therefore, the carbon which is attached to carbonyl carbon is named as alpha (2nd position) and carbon next to alpha carbon is named as beta carbon (3rd position). Hence, we add chlorine atom on 2nd carbon next to carbonyl group and methyl group at 3rd carbon with respect to carbonyl carbon.

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Explanation:

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\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%

If you have 250 mL of a solution that is 44.8 % v/v,

\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}

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