<h3>Answer:</h3>
The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).
<h3>Explanation:</h3>
Alkynes like Alkenes undergo <em>Electrophillic Addition Reactions</em>. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying <em>Markovnikov's rule</em> (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms <em>2-bromo-4,5,5-trimethylhex-1-ene</em>. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce <em>5,5-dibromo-2,2,3-trimethylhexane</em>.
The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.
The theoretical yield of I2 in the reaction would be 0.23 g
<h3>Theoretical yield</h3>
This refers to the stoichiometric yield of a reaction.
From the equation of the reaction:
Ca(IO3)2 + 10 KI + 12 HCl → 6 I2 + CaCl2 + 10 KCl + 6 H2O
The mole ratio of Ca(IO3)2 and I2 is 1: 6
Mole of 15.00 mL, 0.0100 M Ca(IO3)2 = 15/1000 x 0.0100
= 0.00015 mole
Equivalent mole of I2 = 0.00015 x 6
= 0.009 mole
mass of 0.0009 I2 = 0.0009 x 253.809
= 0.23 g
More on stoichiometric calculations can be found here: brainly.com/question/6907332
Answer:
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Explanation:
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Explanation:
