A large activation energy is required to break the double bonds of unsaturated hydrocarbons.
<h3>What are Unsaturated hydrocarbons ?</h3>
Unsaturated hydrocarbons are defined as the hydrocarbons in which double or triple bonds are present between two adjacent carbon atoms. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is CnH2n and CnH2n-2
- In unsaturated hydrocarbons, more number of bonds are formed, thus the bond strength of the bonds formed will be more because the orbitals come closer to each other.
- As, bond strength of unsaturated hydrocarbons are more. So, more energy will be required to break the bond between them.
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Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
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<span> ester of Ethanol and Ethanoic Acid is Ethyl Ethanoate. </span>
<span><span><span><span><span><span><span><span>C</span></span></span><span><span><span>2</span></span></span></span><span><span><span><span>H</span></span></span><span><span><span>5</span></span></span></span><span><span>O</span></span><span><span>H</span></span><span><span>(</span></span><span><span>l</span></span><span><span>)</span></span><span><span>+</span></span><span><span>C</span></span><span><span><span><span>H</span></span></span><span><span><span>3</span></span></span></span><span><span>C</span></span><span><span>O</span></span><span><span>O</span></span><span><span>H</span></span><span><span>(</span></span><span><span>l</span></span><span><span>)</span></span><span><span><span><span><span><span><span><span><span>c</span></span><span><span>o</span></span><span><span>n</span></span><span><span>c</span></span><span><span>.</span></span><span><span><span><span>H</span></span></span><span><span><span>2</span></span></span></span><span><span>S</span></span><span><span><span><span>O</span></span></span><span><span><span>4</span></span></span></span><span><span><span><span>/</span></span></span></span><span><span>w</span></span><span><span>a</span></span><span><span>r</span></span><span><span>m</span></span></span></span><span /></span></span></span><span><span><span><span>−</span><span>−−−−−−−−−−</span><span>→</span></span></span></span></span></span><span><span>C</span></span><span><span><span><span>H</span></span></span><span><span><span>3</span></span></span></span><span><span>C</span></span><span><span>O</span></span><span><span>O</span></span><span><span>C</span></span><span><span><span><span>H</span></span></span><span><span><span>2</span></span></span></span><span><span>C</span></span><span><span><span><span>H</span></span></span><span><span><span>3</span></span></span></span><span><span>(</span></span><span><span>a</span></span><span><span>q</span></span><span><span>)</span></span><span><span>+</span></span><span><span><span><span>H</span></span></span><span><span><span>2</span></span></span></span><span><span>O</span></span><span><span>(</span></span><span><span>l</span></span><span><span>)</span></span></span></span><span>C2H5OH(l)+CH3COOH(l)→conc.H2SO4/warmCH3COOCH2CH3(aq)+H2O(l)</span></span></span>
<span><span><span><span><span><span>Condition: Warm con. reactants with conc.</span></span></span></span><span>Condition: Warm con. reactants with conc.</span></span></span><span><span><span><span><span><span><span><span>H</span></span></span><span><span><span>2</span></span></span></span><span><span>S</span></span><span><span><span><span>O</span></span></span><span><span><span>4</span></span></span></span></span></span><span>H2SO4</span></span></span>
Answer:
the volume is 12.79ml
Explanation:
12.8+12.7+12.78+12.88=51.16
51.16/4
=12.79
