Need answer immediately. For college homework. Organic Chemistry.

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Provide a structure for the following compound: C6H16N2; IR: 3281 cm–1; 1H NMR: δ 1.1 (8H, t, J = 7 Hz), δ 2.66 (4H, q, J = 7 Hz), δ 2.83 (4H, s). (Hint: The triplet at δ 1.1 conceals another broad resonance that contributes to the integral.)
THIS IS THE ANSWER: CH3-CH2-NH-CH2-CH2-NH-CH2-CH3
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Answer:
the answer is below.
Explanation:
CH3-CH2-NH-CH2-CH2-NH-CH2-CH3