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Nataly_w [17]
3 years ago
13

The product of the nitration reaction will have a nitro group at which position with respect to the methyl group? Group of answe

r choices ortho and para at a 50:50 ratio mostly ortho mostly para meta
Chemistry
1 answer:
zimovet [89]3 years ago
4 0

Answer:

Mostly Para

Explanation:

First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).

Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.

Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.

Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.

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If 50 g of lead (of specific heat 0.11 kcal/kg ∙ C°) at 100°C is put into 75 g of water (of specific heat 1.0 kcal/kg ∙ C°) at 0
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Answer: The final temperature is 279.8K

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heat_{absorbed}=heat_{released}

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Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of lead = 50 g

m_2 = mass of water = 75 g

T_{final} = final temperature = ?

T_1 = temperature of lead = 100^oC=373K

T_2 = temperature of water = 0^oC=273K

c_1 = specific heat of lead = 0.11kcal/kg^0C

c_2 = specific heat of water= 1.0kcal/kg^0C

Now put all the given values in equation (1), we get

50\times 0.11\times (T_{final}-373)=-[75\times 1.0\times (T_{final}-273)]

T_{final}=279.8K

Therefore, the final temperature of the mixture will be 279.8 K.

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