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Determine the molecular formula, functional groups, reaction with aqueous bromine and sodium carbonate, polymerization, and polymer type for the given organic structure.

Chemistry

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Answer

b(i)

C4H6O3\text{C}_{4}\text{H}_{6}\text{O}_{3}

Step-by-step solution

2 steps

Part b(i)· 2 steps

Step 1: Count the number of carbon atoms.

There are 4 carbon atoms in the structure.

  1. Count the number of hydrogen atoms.

    The -COOH group has 1 H.
    The left carbon of the C=C bond has 1 H.
    The right carbon of the C=C bond has 1 H.
    The -CH2_{2}OH group has 2 H on the carbon and 1 H on the oxygen.
    Total H = 1+1+1+2+1=61 + 1 + 1 + 2 + 1 = 6 hydrogen atoms.

  2. Count the number of oxygen atoms.

    The -COOH group has 2 O atoms.
    The -CH2_{2}OH group has 1 O atom.
    Total O = 2+1=32 + 1 = 3 oxygen atoms.

    The molecular formula is C4H6O3\boxed{\text{C}_{4}\text{H}_{6}\text{O}_{3}}.

Part b(ii)

  • -C=C-: Alkenes
  • -OH: Alcohols
  • -COOH: Carboxylic acids

Part b(iii)

  • aqueous bromine: The reddish-brown color of aqueous bromine decolorizes due to the presence of the C=C double bond.
  • aqueous sodium carbonate: Effervescence (fizzing) is observed as carbon dioxide gas is produced due to the reaction of the carboxylic acid group with the carbonate.

Part b(iv)

The addition polymer forms by breaking the C=C double bond.

[COOHH∣∣—C—C—∣∣HCH2OH]n\begin{bmatrix} \text{COOH} & \text{H} \\ | & | \\ \text{—C} & \text{—C—} \\ | & | \\ \text{H} & \text{CH}_{2}\text{OH} \end{bmatrix}_{n}

Part b(v)

The type of condensation polymer formed from a dicarboxylic acid and a diol is a polyester.

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