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Marrrta [24]
4 years ago
5

What is the role of a strong acid catalyst in an elimination reaction?

Chemistry
1 answer:
Ivan4 years ago
8 0

Explanation:

An elimination reaction is a type of organic reaction in which two substituents are removed from a molecule in either a one or two-step mechanism. The one-step mechanism is known as the E2 reaction, and the two-step mechanism is known as the E1 reaction.

Another definiton is;

Elimination reaction, any of a class of organic chemical reactions in which a pair of atoms or groups of atoms are removed from a molecule, usually through the action of acids, bases, or metals and, in some cases, by heating to a high temperature. It is the principal process by which organic compounds containing only single carbon-carbon bonds (saturated compounds) are transformed to compounds containing double or triple carbon-carbon bonds (unsaturated compounds).

Strong Acids play two key roles in elimination reactions:

1. Without the acid catalyst, the reaction is painfully slow.

2. Acid greatly facilitates elimination of the leaving group.

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A neutral solution has a(n) _______ amount of hydroxide ions compared to hydrogen ions.
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1.31 times 10^-22 / 6.6262 times 10^-34
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Determine if the bond between each pair of atoms is pure covalent, polar covalent, or ionic. drag the appropriate items to their
dsp73

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Pure Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Br and Br,

                    E.N of Bromine      =   2.96

                    E.N of Bromine      =   2.96

                                                   ________

                    E.N Difference             0.00         (Non Polar/Pure Covalent)

 

For N and O,

                    E.N of Oxygen      =   3.44

                    E.N of Nitrogen     =   3.04

                                                   ________

                    E.N Difference             0.40           (Non Polar/Pure Covalent)

 

For P and H,

                    E.N of Hydrogen       =   2.20

                    E.N of Phosphorous  =   2.19

                                                              ________

                    E.N Difference                  0.01          (Non Polar/Pure Covalent)

 

For K and O,

                    E.N of Oxygen          =   3.44

                    E.N of Potassium      =   0.82

                                                   ________

                    E.N Difference                2.62              (Ionic)

6 0
4 years ago
Some magnesium powder was mixed with some copper( II)oxide and heated strongly. there was a victorious reaction producing a lot
-BARSIC- [3]

Reaction:

\mathrm{Mg \: + \: CuO \rightarrow Cu\: + \: MgO}

Magnesium is a stronger reducing agent than copper and is thus able to reduce copper(II) oxide.

Products of the reaction: Magnesium oxide and metallic copper.

4 0
3 years ago
Read 2 more answers
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